Suppose we wanted to evaluate the double integral $S = \iint_D 4x + 6y \, dx \, dy$ by first applying a change of variables from $D$ to $R$ : $\begin{aligned} x &= X_1(u, v) = \dfrac{u}{4} + \dfrac{v}{16} \\ \\ y &= X_2(u, v) = \dfrac{u}{3} + \dfrac{v}{6} \end{aligned}$ What is $S$ under the change of variables? If you know an expression within absolute value is non-negative, do not use absolute value at all. $S = \iint_R $ $du \, dv$
Solution: If we have a transformation $\bold{X} : R \to D$, then we can rewrite an integral under the change of variables: $ \iint_D f(x, y) \, dA = \iint_R f(\bold{X}(u, v)) | J(\bold{X}) | \, du \, dv$ First we need to find the absolute value of the Jacobian, $|J(\bold{X})|$. $\begin{aligned} |J(\bold{X})| &= \left| \det \begin{pmatrix} \dfrac{\partial X_1}{\partial u} & \dfrac{\partial X_1}{\partial v} \\ \\ \dfrac{\partial X_2}{\partial u} & \dfrac{\partial X_2}{\partial v} \end{pmatrix} \right| \\ \\ &= \left| \det \begin{pmatrix} \dfrac{1}{4} & \dfrac{1}{16} \\ \\ \dfrac{1}{3} & \dfrac{1}{6} \end{pmatrix} \right| \\ \\ &= \left| \dfrac{1}{24} - \dfrac{1}{48} \right| \\ \\ &= \left| \dfrac{1}{48} \right| \\ \\ &= \dfrac{1}{48} \end{aligned}$ Now we substitute $u$ and $v$ in $f(x, y)$. $\begin{aligned} f(x, y) &= f(\bold{X}(u, v)) \\ \\ &= 4 X_1(u, v) + 6 X_2(u, v) \\ \\ &= 4 \left( \dfrac{u}{4} + \dfrac{v}{16} \right) + 6 \left( \dfrac{u}{3} + \dfrac{v}{6} \right) \\ \\ &= u + \dfrac{v}{4} + 2u + v \\ \\ &= 3u + \dfrac{5v}{4} \end{aligned}$ Putting everything together, we get the integral under the change of variables: $ \iint_R \dfrac{u}{16} + \dfrac{5v}{192} \, du \, dv$